The derivative of s i n − 1 ( 2 x 1 + x 2 ) with respect to t a n − 1 ( 2 x 1 − x 2 ) is Options: A . &nbsp0 B . &nbsp1 C . &nbsp11−x2 D . &nbsp11+x2

Answer: Option B : B L e t u = s i n − 1 ( 2 x 1 + x 2 ) = 2 t a n − 1 x ∴ d u d x = 2 1 + x 2 a n d v = t a n − 1 ( 2 x 1 + x 2 ) = 2 t a n − 1 x ∴ d v d x = 2 1 + x 2 ∴ d u d v = ( d u d x ) ( d v d x ) = 1

The derivative of sin−1(2x1+x2) with respect to tan−1(2x1−x

Question

The derivative of

s i n^{- 1} (\frac{2 x}{1 + x^{2}})

with respect to

t a n^{- 1} (\frac{2 x}{1 - x^{2}})

Options:

A . 0

B . 1

C . 11−x2

D . 11+x2

Answer: Option B
:
B

L e t u = s i n^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 t a n^{- 1} x

∴ \frac{d u}{d x} = \frac{2}{1 + x^{2}}

a n d v = t a n^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 t a n^{- 1} x

∴ \frac{d v}{d x} = \frac{2}{1 + x^{2}}

∴ \frac{d u}{d v} = \frac{(\frac{d u}{d x})}{(\frac{d v}{d x})} = 1

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